For the cubic equation $x^3-3x-1=0$, you can check it has three real roots by calculus knowledge. Let them be $x_1<x_2<x_3$. Now prove that$$x_3^2-x_2^2=x_3-x_1.$$ It is not difficult to check that:$$x_1=2\cos 140 ^{\circ},x_2=2\cos 100 ^{\circ},x_3=2\cos 20 ^{\circ}.$$(Here use the formula: $\cos 3x=4\cos^3x-3\cos x.$ )
And then we can check:$$x_3^2-x_2^2=x_3-x_1.$$I just want to know how to get: $x_1=2\cos 140 ^{\circ},x_2=2\cos 100 ^{\circ},x_3=2\cos 20 ^{\circ}.$ Is there some trick. Any help and hint will welcome.
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$\begingroup$Using the formula $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$ is actually quite useful, but first we have to do $x=2y$, then we get$$8y^3 - 6y - 1 = 0$$Now we have the 4:-3 ratio and we can substitute $y=\cos(\theta)$ to get, from the above formula,$$2\cos(3\theta) = 1$$and from there we can get cosine form of the roots.
$\endgroup$ $\begingroup$Since the roots are of the form $\cos(\theta)$, make the substitution $x=t+t^{-1}$ and expand to get
$$P(t+t^{-1})=\frac {t^6-t^3+1}{t^3}=\frac {t^9+1}{t^3(t^3+1)}$$Set the above expression equal to zero and it's easy to see that$$t^9=-1$$Let $t=\cos\theta+i\sin\theta$ and expand to see that$$\theta=\frac {\pi(2k+1)}{9}$$where $k=0,2,3,\ldots$. Hence$$x=t+t^{-1}=2\cos\left(\frac {\pi(2k+1)}9\right)$$
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